Simplify; express your answer in exponential form. Assume $y\neq 0, r\neq 0$. $\dfrac{{(y^{2}r^{4})^{-1}}}{{(y^{-1}r^{4})^{-4}}}$
Answer: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(y^{2}r^{4})^{-1} = (y^{2})^{-1}(r^{4})^{-1}}$ On the left, we have ${y^{2}}$ to the exponent ${-1}$ . Now ${2 \times -1 = -2}$ , so ${(y^{2})^{-1} = y^{-2}}$ Apply the ideas above to simplify the equation. $\dfrac{{(y^{2}r^{4})^{-1}}}{{(y^{-1}r^{4})^{-4}}} = \dfrac{{y^{-2}r^{-4}}}{{y^{4}r^{-16}}}$ Break up the equation by variable and simplify. $\dfrac{{y^{-2}r^{-4}}}{{y^{4}r^{-16}}} = \dfrac{{y^{-2}}}{{y^{4}}} \cdot \dfrac{{r^{-4}}}{{r^{-16}}} = y^{{-2} - {4}} \cdot r^{{-4} - {(-16)}} = y^{-6}r^{12}$